Timeout with withTimeout in Kotlin - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to run a block with a timeout of 1000 milliseconds.

Kotlin

withTimeout([1]) {
    println("Running task")
}

Drag options to blanks, or click blank then click option'

A100

B5000

C1000

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using seconds instead of milliseconds.

Passing zero which means no timeout.

The withTimeout function takes the timeout duration in milliseconds. Here, 1000 means 1 second.

2fill in blank

medium

Complete the code to catch the timeout exception thrown by withTimeout.

Kotlin

try {
    withTimeout(500) {
        Thread.sleep(1000)
    }
} catch ([1]) {
    println("Timeout occurred")
}

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ATimeoutCancellationException

BException()

CInterruptedException

DCancellationException

Attempts:

3 left

💡 Hint

Common Mistakes

Catching a generic Exception instead of the specific timeout exception.

Using InterruptedException which is unrelated here.

The withTimeout function throws TimeoutCancellationException when the timeout expires.

3fill in blank

hard

Fix the error in the code to properly use withTimeout in a suspend function.

Kotlin

suspend fun runTask() {
    withTimeout([1]) {
        delay(2000)
    }
}

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Adelay(1000)

B1000L

C"1000"

D1000

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a string instead of an integer.

Passing a Long literal which is allowed but not necessary here.

Passing a function call instead of a number.

The timeout duration must be an integer number of milliseconds. Using 1000 is correct. Quotes make it a string which is invalid.

4fill in blank

hard

Fill both blanks to create a map of words to their lengths only if length is greater than 3.

Kotlin

val words = listOf("apple", "bat", "carrot", "dog")
val lengths = words.filter { word -> [2] }.associate { word -> word to [1] }

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Aword.length

Blen(word)

Cword.length > 3

Dlen(word) > 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using Python syntax like len(word).

Using incorrect condition syntax.

In Kotlin, word.length gets the length of the string. The condition word.length > 3 filters words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a map of uppercase words to their lengths only if length is less than 5.

Kotlin

val words = listOf("apple", "bat", "carrot", "dog")
val result = words.filter { word -> [3] }.associate { word -> [1] to [2] }

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Aword.uppercase()

Bword.length

Cword.length < 5

Dword.toUpperCase()

Attempts:

3 left

💡 Hint

Common Mistakes

Using deprecated toUpperCase() instead of uppercase().

Using Python syntax like len(word).

Incorrect condition operator.

Use word.uppercase() to convert to uppercase in Kotlin. word.length gets length. The condition word.length < 5 filters words shorter than 5.